When deleting words (ctrl+backspace or ctrl+delete) in IntelliJ it treats whitespace as a word until it hits a newline character. I want to treat every sequence of space/tab and newline[sequences] as one word.
Essentially this is the reverse problem of this question in IntelliJ.
Sadly i did not find any setting to do that so far.
There exists a "Hungry Backspace" action.
Yet there exists no "Hungry Delete" out of the box.
And it does not work for non-whitespace words.
Related
How can I select a XPath that contains a text with both quote & comma character?
I have to find an element if the text contains =Yes, It's OK
My XPath does not save in ranorex tool even though if i put the text inside the double quotes like below
//span[text()="Yes, It's OK"]
So how can I save this xpath that uses "".in Ranorex
your use of double-quotes is correct. if
//span[text()="Yes, It's OK"]
doesn't match, it might be because the xpath engine has a lowercase bug (i have encountered PHP+DOMXPath+libxml2 systems where it would only match lowercase text even for text with uppercase characters, never quite figured out the problem there), or it might be because the text has hidden whitespace you're not aware of, maybe it actually has a hard-to-spot whitespace at the start or the end? anyway maybe contains() will work:
//span[contains(text(),"Yes, It's OK"]
or.. if it has the lowercase-issue i have encountered in the wild once before, maybe this will work:
//span[contains(text(),"yes, it's ok"]
(that really shouldn't be required, but i have been in that situation once before, and that was years ago, and that was... probably php5)
Hello I want to know any trick or shortcut by which one can know which is the longest file in project.
i.e which file has the longest lines of code.Is there any shortcut or plugin available?
I believe the OP was asking about the length of file, not the length of single line. You can try with such iteration:
(.*\n){100,}
(.*\n){1000,}
(.*\n){10000,}
Although this is kind of hacky it still works.
You can search your whole project using the regex repetition pattern. Just right-click your project folder in the project structure view and choose "Find in path...". Be sure to check "Regex" in the search window that appears.
So you'll start out and match any line with any length in your project
^.$
(If you're not familiar with regex: ^ and $ are used to denote the beginning and end of a line and . matches any character)
Then you gradually increase the number of matched repetitions
^.{1,}$
^.{10,}$
^.{100,}$
^.{1000,}$
(You use {start, end} to indicate to interval of repetitions. If you leave end blank it will match anything from start)
Using this you will soon be left with the longest line(s) in your project.
As I said it's kinda hacky but it's also quick and works if you don't have to automate the task.
Hope this helps you!
All.
I am used to programming VBA in Excel, but am new to the structures in Word.
I am working through a library of text files to update them. Many of them are either OCR documents, or were manually entered.
Each has a recurring pattern, the most common of which is unnecessary carriage returns.
For example, I am looking at several text files where there is a double return after each line. A search and replace of all double carriage returns removes all paragraph distinctions.
However, each line is approximately 30 characters long, and if I manually perform the following logic, it gives me a functional document.
If there is a double carriage return after 30+ characters, I replace them with a space.
If there were less than 30 characters prior to the double return, I replace them with a single return.
Can anyone help me with some rudimentary code that would help me get started on that? I could then modify it for each "pattern" of text documents I have.
e.g.
In this case, there are more than
thirty characters per line. And I
will keep going to illustrate this
example.
This would be a new paragraph, and
would be separated by another of
the single returns.
I want code that would return:
In this case, there are more than thirty character returns. And I will keep going to illustrate this example.
This would be a new paragraph, and would be separated by another of the single returns.
Let me know if anyone can throw something out that I can play with!
You can do this without code (which RegEx requires), simply using Word's own wildcard Find/Replace tools, where:
Find = ([!^13]{30,})[^13]{1,}
Replace = \1^32
and, to clean up the residual multi-paragraph breaks:
Find = [^13]{2,}
Replace = ^p
You could, of course, record the above as a macro...
Here is a RegEx that might work for you:
(\n\n)(?<!\.(\n\n))
The substitution is just a plain space, you can try it out (and modify / tweak it) here: https://regex101.com/r/zG9GPw/4
This 'pattern' tells the RegEx engine to look for the newline character \n which occurs x2 like this \n\n (worth noting this is from your question and might be different in your files, e.g. could be \r\n) and it assumes that a valid line break will be proceeded by a full stop: \..
In RegEx the full stop symbol is a single character wild card so it needs to be escaped with the '\' (n and r are normal characters, escaping them tells the RegEx engine they represent newline and return characters).
So... the expression is looking for a group of x2 newline characters but then uses a negative look-behind to exclude any matches where the previous character was a full stop.
Anyway, it's all explained on the site:
Here is how you could do a RegEx find and replace using NotePad++ (I'm not sure if it comes with RegEx or if a plugin is needed, either way it is easy). But you can set a location, filters (to target specific file types), and other options (such as search in sub-directories).
Other than that, as #MacroPod pointed out you could also do this with MS Word, document by document, not using any code :)
Probably quite a niche question, but I believe in the power of a big community: Is it possible to set up jEdit in way, that it automatically inserts a comment character (//, #, ... depending on the edit mode) at the beginning of a new line, if the line before the wrap was a comment?
Sample:
# This is a comment spanning multiple lines. If I continue to type here, it
# wraps around automatically, but I have to manually add a `#` to each line.
If I continue to type after the . the third line should start with the # automatically. I searched in the plugin repository but could not find anything related.
Background: jEdit has the concepct of soft and hard wrap. While soft wrap only breaks lines visually at a character limit, it does not insert line breaks in the file. Hard wrap on the other hand inserts \n into the file at the desired character count.
This is not exactly what you want: I use the macros Enter_with_Prefix.bsh to automatically insert the prefix (e.g., #, //) at the beginning of the new line.
Description copied from Enter_with_Prefix.bsh:
Enter_with_Prefix.bsh - a Beanshell macro for jEdit
that starts a new line continuing any recognized
sequence that started the previous. For example,
if the previous line beings with "1." the next will
be prefixed with "2.". It supports alpha lists (a., b., etc...),
bullet lists (+, =, *, etc..), comments, Javadocs,
Java import statements, e-mail replies (>, |, :),
and is easy to extend with new sequence types. Suggested
shortcut for this macro is S+ENTER (SHIFT+ENTER).
How to search word start \word in vim. I can do it using the find menu. Is there any other short cut for this?
Try:
/\\word
in command mode.
You can search for most anything in your document using regular expressions. From normal mode, type '/' and then start typing your regular expression, and then press enter. '\<' would match the beginning of a word, so
/\<foo
would match the string 'foo' but only where it is at the beginning of a word (preceded by whitespace in most cases).
You can search for the backslash character by escaping it with a backslash, so:
/\<\\foo
Would find the pattern '\foo' at the beginning of a word.
Not directly relevant (/\\word is the the correct solution, and nothing here changes that), but for your information:
:h magic
If you are for a pattern with many characters with special meaning to regexes, you may find "nomagic" and "very nomagic" mode useful.
/\V^.$
will search for the literal string ^.$, instead of "lines of exactly one character" (\v "very magic" and the default \m "magic" modes) or "lines of exactly one period" (\M "nomagic" mode).
The reason searching for something including "\" is different is because "\" is a special character and needs to be escaped (prepended with a backslash)
Similarly, to search for "$100", which includes the special character "$":
Press /
Type \$100
Press return
To search for "abc", which doesn't include a special character:
Press /
Type abc
Press return